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Random walk versus random line
Joel de Coninck, Francois Dunlop, Thierry Huillet
To cite this version:
Joel de Coninck, Francois Dunlop, Thierry Huillet. Random walk versus random line. Physica A:
Statistical and Theoretical Physics, Elsevier, 2009, 388, pp.4034-4040. �10.1016/j.physa.2009.06.030�.
�hal-00375957�
Random walk versus random line
Jo¨el De Coninck
(1), Fran¸cois Dunlop
(2), Thierry Huillet
(2)Abstract: We consider random walks X
nin Z
+, obeying a detailed balance condition, with a weak drift towards the origin when X
nր ∞. We reconsider the equivalence in law between a random walk bridge and a 1+1 dimensional Solid-On-Solid bridge with a corresponding Hamiltonian. Phase diagrams are discussed in terms of recurrence versus wetting. A drift −δX
n−1+ O(X
n−2) of the random walk yields a Solid-On-Solid potential with an attractive well at the origin and a repulsive tail
δ(2+δ)8X
n−2+ O(X
n−3) at infinity, showing complete wetting for δ ≤ 1 and critical partial wetting for δ > 1.
KEYWORDS: Random walk, recurrence, SOS model, pinning, wetting AMS subject classification: 60J10, 82B41
1. Introduction
We consider a random walk on Z or Z
+as defined by transition probabilities P (X
n+1|X
n), so that the probability distribution of a random walk bridge of length N is
P (X
1, . . . , X
N−1|X
0= X
N= 0) =
N−1
Y
n=0
P (X
n+1|X
n) .
P (X
N= 0|X
0= 0) (1.1)
We consider a random line making a bridge of length N , in the form of a Solid-On-Solid model, as defined by a probability distribution of the form
P (X
1, . . . , X
N−1|X
0= X
N= 0) = Z
N−1N−1
Y
n=0
e
−W(Xn,Xn+1)N
Y
n=1
e
−V(Xn)(1.2)
with W (X, Y ) = W (Y, X ) for all X, Y , and Z
Nthe partition function normalising the probability.
We address the question of translating P (X
n+1|X
n) into W (X, Y ) and V (X ), and conversely, and transferring the information about transience / null recurrence / positive recurrence of the walk to complete wetting / partial wetting of the SOS model, and back.
This question is related to the Hamiltonian on random walk trajectories in Ferrari-Mart´ınez [FM].
(1) Centre de Recherche en Mod´elisation Mol´eculaire, Universit´e de Mons-Hainaut, 20 Place du Parc, 7000 Mons, Belgium. Email: Joel.De.Coninck@crmm.umh.ac.be
(2) Laboratoire de Physique Th´eorique et Mod´elisation (CNRS - UMR 8089), Universit´e de Cergy- Pontoise, 95302 Cergy-Pontoise, France. Email: Francois.Dunlop@u-cergy.fr, Thierry.Huillet@u-cergy.fr
We assume that the walk obeys the detailed balance condition with respect to a measure on Z , not necessarily normalisable, which we write as exp(−U (X)), so that
P (X
n+1|X
n) = e
U(Xn)−U(Xn+1)P (X
n|X
n+1)
= e
12U(Xn)P (X
n+1|X
n) P (X
n|X
n+1)
12e
−12U(Xn+1)≡ e
12U(Xn)e
−W(Xn,Xn+1)e
−12U(Xn+1)(1.3)
which defines W (X, Y ) from P (Y |X ). The probability of a random walk bridge may now be written as
P (X
1, . . . , X
N−1|X
N= X
0= 0) = Z
N−1N−1
Y
n=0
e
−W(Xn,Xn+1)(1.4) with Z
N= P (X
N= 0|X
0= 0), which is of the form (1.2). The detailed balance condition was used, but the formula implied by (1.3) for the resulting SOS interaction W does not re- quire the knowledge of the invariant measure exp(−U (X )). The interaction W (X
n, X
n+1) typically contains a part of the form (V (X
n) + V (X
n+1))/2, which may be split from W . Conversely, given a SOS probability distribution of the form (1.2), where we let W absorb V like in (1.4), we look for a set of random walk probability transitions of the form
P (X
n+1|X
n) = e
−W(Xn,Xn+1)−12U(Xn+1)+12U(Xn)Z(X
n) (1.5)
These would lead to
P (X
1, . . . , X
N−1|X
0= X
N= 0) =
N−1
Y
n=0
e
−W(Xn,Xn+1)Z(X
n) (1.6)
which agrees with (1.4) only if Z = const., which requires exp(−
12U ) to be an eigenvector of the symmetric kernel exp(−W (X, Y )):
X
X
e
−12U(X)e
−W(X,Y)= ρ e
−12U(Y)(1.7) The Perron-Frobenius theorem [S] indicates that (1.7) should have a solution (ρ, U ). In any case, (1.7) is equivalent to exp(−U ) being a left-eigenvector of the (non-symmetric) kernel (1.5) with Z = const.:
X
X
e
−U(X)e
−W(X,Y)−12U(Y)+12U(X)= ρ e
−U(Y)(1.8)
Therefore (1.5) with Z = const. and U obeying (1.7) or (1.8) is an answer to formulating
an SOS random line with probability (1.2), written as (1.4), in terms of a random walk.
However, it does require the knowledge of the measure exp(−U (X )), with respect to which the walk will obey the detailed balance condition. This is related to the transfer matrix solution of the 1+1 dimensional SOS models of wetting derived in the early eighties [AD, Bu, C, CW, LH, VL] and further elaborated with path space limit theorems in the late nineties [Bo, DGZ, IY, V] and references therein. Expressing an SOS bridge in terms of a random walk, asymptotically as N → ∞, was used also in the proof of the Wulff shape for SOS models (Theorem 1 in [DDR]).
In the following sections we consider examples, translating from random walk to SOS model, when
P (X
n+1< X
n|X
n) − P (X
n+1> X
n|X
n) ∼ δ X
nas X
n→ ∞ (1.9)
and discuss recurrence versus wetting. Interest into such random walks goes back to Lamperti [L1, L2]. Detailed properties of the random walk are available [DDH, H] in special instances of (1.9), yielding the corresponding properties in the corresponding SOS models. Some of these examples admit constructions for bridges not using the detailed balance formula.
2. Bridge with X
n+1− X
n= ±1: from random walk to random line Let
ϕ : {
12,32,52, . . .} → R
Consider a random walk X
nwith state space Z
+= {0, 1, 2, . . .}, starting at X
0= 0, with transition probabilities
P (X
n+1|X
n) = e
−(Xn+1−Xn)ϕ(Xn+1 +2 Xn)e
−ϕ(Xn+12)+ e
ϕ(Xn−12)when X
n≥ 1 and X
n+1= X
n± 1 (2.1) and reflection at the origin: X
n+1= 1 whenever X
n= 0. Any random walk with transition probabilities p
x= P (X
n+1= x + 1|X
n= x) and q
x= 1 − p
x= P (X
n+1= x − 1|X
n= x) may be written in the form (2.1): take ϕ(
12) arbitrarily, and then solve recursively
ϕ(x +
12) = −ϕ(x −
12) + ln q
xp
x, x ≥ 1 (2.2)
From (2.1) we get
P (X
1, . . . , X
N−1, X
N= 0|X
0= 0) =
N−1
Y
n=0
P (X
n+1|X
n)
=
N
Y
n=1 Xn=0
e
ϕ(12)N
Y
n=1 Xn≥1
1
e
−ϕ(Xn+12)+ e
ϕ(Xn−12)N−1
Y
n=0
1
|Xn+1−Xn|=1= 2
−NN
Y
n=1 Xn=0
2e
ϕ(12)N
Y
n=1 Xn≥1
2
e
−ϕ(Xn+12)+ e
ϕ(Xn−12)N−1
Y
n=0
1
|Xn+1−Xn|=1= 2
−NN
Y
n=1
e
−V(Xn)N−1
Y
n=0
1
|Xn+1−Xn|=1(2.3)
with
V (X) = −
ln 2 + ϕ( 1 2 )
1
X=0+ ln e
−ϕ(X+12)+ e
ϕ(X−12)2 1
X≥1(2.4)
The key point in the computation (2.3), instead of using the detailed balance condition, was the pairing of edge factors, one factor corresponding to going up the edge and the other factor going down the edge, leading to the cancellation of factors from the numerator in (2.1). This exact cancellation is restricted to bridges, and requires the coupling ϕ in (2.1) to be associated with the un-oriented edge {X
n, X
n+1} or to the midpoint (X
n+ X
n+1)/2.
Example (see Fig 1):
ϕ(x) = δ
2x ⇒ V (X) = −(ln 2 + δ) 1
X=0+ ln e
−2X+1δ+ e
2X−1δ2 1
X≥1∼ δ(2 + δ)
8X
2as X → ∞
(2.5)
Such a potential for δ > 0, having short range attraction at the wall and long range repulsion far from the wall, is reminiscent of van der Waals liquids with a positive Hamaker constant [dG, p846]. The 1+1 dimensional SOS model may be considered a crude effective interface model where some dimensions and degrees of freedom have been integrated out in a mean field approximation.
-2 -1.5 -1 -0.5 0 0.5 1
0 1 2 3 4 5
1.2 0.5 -0.2 -1.2
Fig. 1: V(X) as (2.4) with
δ = 1.2, 0.5, −0.2, −1.2
.Example:
ϕ(x) = δ 2x + γ
x
2+ O 1 x
3as x → ∞ (2.6)
Such random walks should have a phase diagram (transience / null recurrence / positive recurrence) independent of γ , and also independent of the behaviour of ϕ for small x.
Hence the corresponding SOS models should have a phase diagram (complete / partial wetting) independent of γ: partial wetting if and only if δ > 1. However, unlike the square well model in the partial wetting regime (cf. next section), the height distribution will not decay exponentially, but as a power law with an exponent depending upon δ [DDH], hence the term “critical partial wetting”.
The behaviour (2.6) implies
V (X) = δ(2 + δ)
8X
2+ O 1 X
3as X → ∞ (2.7)
which indeed is independent of γ .
3. Bridge with X
n+1− X
n= ±1: from random line to random walk
Suppose now that the potential V (X) on Z
+is given and satisfies V (X ) → 0 as X → ∞.
We want to find ϕ : {
12,32,52, . . .} → R such that (2.4) is satisfied up to a constant λ. Let
b
X= e
V(X)+λ, a
X= e
−ϕ(X+12 )(3.1)
Then (2.4) with V (X) + λ instead of V (X ) becomes 2b
0= a
02b
X= a
X+ a
−1X−1, X ≥ 1 (3.2)
whose solution is the continued fraction a
0= 2b
0a
1= 2b
1− 1 2b
0· · ·
a
X= 2b
X− 1
2b
X−1−
2b 1X−2− ··· ··· ··· ··· ··· ··· ··· ···1
··· ··· ··· ··· ··· ··· ···
2b3− 1 2b2− 1
2b1− 1 2b0
(3.3)
acceptable only if a
X> 0 ∀X . Consistency may be verified when (2.3)(2.4), converted into (1.4), obeys (1.7), which takes the form
X
X=Y±1 X≥0
e
−12U(X)−12V(X)−12V(Y)= 2ρ e
−12U(Y), Y ≥ 0 (3.4)
so that (3.2)(3.3)(3.4) have the solution λ = ln ρ and
a
X= e
−12U(X+1)−12V(X+1)+12U(X)+12V(X), X ≥ 0 (3.5) To conclude this section, we give explicitly the random walks corresponding to the SOS model with a square well or a double step potential at the wall:
• For V (X) = v
01
X=0, equations (3.1)(3.2) with ρ = e
λtake the form 2b
0= 2ρe
v0= a
02b
X= 2ρ = a
X+ a
−1X−1, X ≥ 1 (3.6)
— First ansatz: ρ = 1
a
X= (2b
0− 1)X + 2b
0(2b
0− 1)X + 1 > 0 ∀X ⇒ v
0≥ − ln 2 (3.7) a transient walk with
ϕ(x) ∼ − 1
x as x → ∞ (3.8)
compatible with (2.4), δ = −2.
— Second ansatz: a
X= a = const.
ρ = a + a
−12 , a
−2= e
−v0− 1 > 0 ⇒ v
0< 0 (3.9) Both ansatz work when − ln 2 ≤ v
0< 0, corresponding to transient cases. The wetting transition is at v
0= − ln 2.
• For V (X) = v
01
X=0+ v
11
X=1, equations (3.1)(3.2) with ρ = e
λtake the form 2b
0= 2ρe
v0= a
02b
1= 2ρe
v1= a
1+ a
−102b
X= 2ρ = a
X+ a
−1X−1, X ≥ 2
(3.10)
— First ansatz: ρ = 1 a
0= 2b
0a
1= 2b
1− 1
2b
0> 0 a
X= (a
1− 1)X + 1
(a
1− 1)X + 2 − a
1> 0 ∀X ≥ 2 ⇒ a
1= 2b
1− 1 2b
0≥ 1
(3.11)
or
4e
v1≥ 2 + e
−v0(3.12)
a transient walk with ϕ(x) ∼ −
1xas x → ∞. Condition (3.12) coincides with the complete wetting range.
— Second ansatz: a
X= a = const. ∀X ≥ 1 a
0= 2ρe
v0a = 2ρe
v1− 1 2ρe
v0ρ = a + a
−12
(3.13)
Eliminating ρ gives
a
4(e
v1− 1) + a
2(2e
v1− e
−v0− 1) + e
v1= 0 (3.14) giving a suitable solution for v
1≤ 0 and any v
0and also for
v
1≥ 0 , v
0≤ 0 , v
1≤ 2 log cosh v
02 (3.15)
Whatever v
0and v
1, one or the other or both ansatz provides a solution. There is partial wetting if and only if there is a representation with 0 < a < 1, equivalent to
4e
v1< 2 + e
−v0(3.16)
where only the second ansatz gives a solution, in fact one solution if v
1≤ 0 and two
solutions if v
1> 0.
4. Bridge with X
n+1− X
n∈ {−1, 0, +1} , Metropolis algorithm Let
U : Z
+→ R
Consider a random walk X
nwith state space Z
+= {0, 1, 2, . . .}, starting at X
0= 0, with transition probabilities
P (X
n+1|X
n, X
n≥ 1) = 1
Xn+1=Xn±1 12
e
− U(Xn+1)−U(Xn)+
+ 1
Xn+1=Xnh 1 −
12e
− U(Xn+1)−U(Xn)+
−
12e
− U(Xn−1)−U(Xn)+
i
(4.1) and reflection at the origin: X
n+1= 1 whenever X
n= 0. Then
P (X
1, . . . , X
N−1, X
N= 0|X
0= 0) =
N−1
Y
n=0
P (X
n+1|X
n)
=
N−1
Y
n=1 Xn+1 =Xn6=0
h 1 −
12e
− U(Xn+1)−U(Xn)+
−
12e
− U(Xn−1)−U(Xn)+
i .
.
N−1
Y
n=1 Xn+1 =Xn±1
Xn,Xn+16=0
1 2
e
−|U(Xn+1 )−U(Xn)|
2
N
Y
n=1 Xn=0
1
2
e
−(U(0)−U(1))+=
N−1
Y
n=0 Xn+1 =Xn6=0
h 1 −
12e
− U(Xn+1)−U(Xn)+
−
12e
− U(Xn−1)−U(Xn)+
i .
.
N−1
Y
n=0 Xn+1 =Xn±1
1 2
e
−|U(Xn+1 )−U(Xn)|
2
N
Y
n=1 Xn=0
2e
(U(1)−U(0))+= 2
−NN−1
Y
n=0
e
−W(Xn,Xn+1)N
Y
n=1
e
−V(Xn)(4.2) with
V (X ) = −
ln 2 + (U (1) − U (0))
+1
X=0W (X, X ) = − ln h
2 − e
− U(X+1)−U(X)+
− e
− U(X−1)−U(X)+
i
W (X, X + 1) = W (X + 1, X ) = |U (X + 1) − U (X)|
2
(4.3)
where W (X, X) is used only with X ≥ 1. The pairing of edge factors was used, like in
Section 2.
Example: δ ≥ 0 and
U (X ) = δ ln(X + 1) ⇒ V (X) = −(ln 2 + δ ln 2) 1
X=0W (X, X ) = − ln
1 − X + 1 X + 2
δW (X + 1, X ) = W (X, X + 1) = δ
2 ln X + 2 X + 1
(4.4)
Instead of reflection at the origin, let us now choose the full Metropolis algorithm, including at the wall:
P (X
n+1|X
n= 0) =
12e
−(U(1)−U(0))+1
Xn+1=1+ (1 −
12e
−(U(1)−U(0))+) 1
Xn+1=0(4.5) Then
P (X
1, . . . , X
N−1, X
N= 0|X
0= 0) =
N−1
Y
n=0
P (X
n+1|X
n)
=
N−1
Y
n=0 Xn+1 =Xn6=0
h
1 −
12e
− U(Xn+1)−U(Xn)+
−
12e
− U(Xn−1)−U(Xn)+
i .
.
N−1
Y
n=0 Xn+1 =Xn=0
h 1 −
12e
− U(1)−U(0)+
i
N−1Y
n=0 Xn+1 =Xn±1
1
2
e
−|U(Xn+1 )2−U(Xn)|= 2
−NN−1
Y
n=0
e
−W(Xn,Xn+1)(4.6) with
W (X, X + 1) = W (X + 1, X) = |U (X + 1) − U (X )|
2 W (X, X ) = − ln h
2 − e
− U(X+1)−U(X)+
− e
− U(X−1)−U(X)+
i
except : W (0, 0) = − ln h
2 − e
− U(1)−U(0)+
i
(4.7) Example: δ ≥ 0 and
U (X) = δ ln(X + 1) ⇒ W (0, 0) = − ln 2 − 2
−δ(4.8) and the other values same as first Metropolis example.
Remark: The factor 1/2 in (4.1) could be replaced by any number between 0 and 1/2.
5. Random walk with X
n+1− X
n∈ Z , Metropolis algorithm Let exp(−W
0(X, Y ) be a symmetric probability kernel in Z × Z ,
W
0(X, Y ) = W
0(Y, X ) , X
Y∈Z
e
−W0(X,Y)= 1 (5.1)
and
U : Z
+→ { R ∪ {+∞}} , with : X < 0 ⇒ U (X ) = +∞ (5.2) Consider a random walk X
nwith state space Z
+= {0, 1, 2, . . .}, starting at X
0= 0, with transition probabilities
P (X
n+1|X
n) = e
−W0(Xn+1,Xn)− U(Xn+1)−U(Xn)+
if X
n+16= X
nP (X
n+1= X
n|X
n) = 1 − X
Y6=Xn
P (Y |X
n) (5.3)
Then, proceeding as in Section 1, we get (1.4) with
W (X, Y ) = W (Y, X ) = W
0(X, Y ) + |U (Y ) − U (X )|
2 if Y 6= X
W (X, X ) = − ln
1 − X
Y∈Z
e
−W0(X,Y)− U(Y)−U(X)+